在字符的第一次和最后一次出现时拆分？

Question

I have a list of strings as such (amount, address, payment):

"44.53 54 orchard rd Cash"
"32.34 600 sprout brook lane Card"

I am just trying to get the address from each string. It seems to me the best way to go about this would be to split at the first and last occurrence of a space. Is there any way to do this?

Answer 1

Python split function is defined like this: str.split(sep=None, maxsplit=-1) .

Similarly, there is str.rsplit(sep=None, maxsplit=-1) .

This means that you can split off just the beginning and the ending:

>>> s = "44.53 54 orchard rd Cash"
>>> s.split(maxsplit=1)
['44.53', '54 orchard rd Cash']
>>> s.rsplit(maxsplit=1)
['44.53 54 orchard rd', 'Cash']

Then, to simply split the string into 3, you can write a simple function:

>>> def purchase_parts(purchase):
...     lsplit = purchase.split(maxsplit=1)
...     rsplit = lsplit[1].rsplit(maxsplit=1)
...     return (lsplit[0], rsplit[0], rsplit[1])
... 
>>> purchase_parts("44.53 54 orchard rd Cash")
('44.53', '54 orchard rd', 'Cash')
>>> purchase_parts("32.34 600 sprout brook lane Card")
('32.34', '600 sprout brook lane', 'Card')

Still, I would suggest to switch to separated value list, because then you can just split using that separator, but also directly support importing/exporting of csv format (comma separated values) files.

Manual solution:

>>> [p.strip() for p in "32.34, 600 sprout brook lane, Card".split(',')]
['32.34', '600 sprout brook lane', 'Card']

Answer 2

You could potentially do something like:

line = "44.53 54 orchard rd Cash"
line_parts = line.split(" ")
address = " ".join(line_parts[1:-1])

It's a bit untidy and definitely brittle to changes in line format, but would do the job.

Answer 3

You can use your method, splitting at the first and last spaces, but you need to join back the rest (in the middle):

def get_address(s):
    s = s.split()
    return ' '.join(s[1:-1])
    # s[1:-1] will remove the first (amount) and the last (payment) values
    # ' '.join will then put back the spaces that were removed from the address by s.split

Input:

print(get_address("44.53 54 orchard rd Cash"))
print(get_address("32.34 600 sprout brook lane Cash"))

Output:

54 orchard rd
600 sprout brook lane

Answer 4

You could also use a regular expression to be a bit more flexible and robust. Here, the first two \\d+ elements say that you must at first have two digits separated by a dot, then a space, then your address as returned result (in parenthesis () ) consisting of any characters ( \\w ) or ( [] ) whitespace characters ( \\W ) until a space and another sequence of characters ( \\w+ ).

import re

addresses = [
    "44.53 54 orchard rd Cash",
    "32.34 600 sprout brook lane Card"
]

addresses = [re.findall(r'\d+\.\d+ ([\w\W]+) \w+', address)[0] for address in addresses]
print(addresses)  # ['54 orchard rd', '600 sprout brook lane']

Answer 5

You could get the first and last using unpacking and the reassemble then rest to form the address:

amount,*rest,payment = s.split()
address = " ".join(rest)